integral from 0 to 1 of-2sin((2-2x)^2)

解

\int_{0}^{1} - 2 sin ((2 - 2 x)^{2}) d x

- \frac{π}{2} § (2 \frac{2}{π}) + \frac{π}{2} § (0 \cdot \frac{2}{π})

解答ステップ

\int_{0}^{1} - 2 sin ((2 - 2 x)^{2}) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int_{0}^{1} sin ((2 - 2 x)^{2}) d x

u置換積分法を適用する

= - 2 \cdot \int_{2}^{0} - \frac{1}{2} sin (u^{2}) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - 2 (- \int_{0}^{2} - \frac{1}{2} sin (u^{2}) d u)

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 (- (- \frac{1}{2} \cdot \int_{0}^{2} sin (u^{2}) d u))

非初等積分を使用する:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= - 2 - - \frac{1}{2} [\frac{π}{2} § (\frac{2}{π} u)]_{0}^{2}

簡素化

- 2 - - \frac{1}{2} [\frac{π}{2} § (\frac{2}{π} u)]_{0}^{2} : - [\frac{π}{2} § (\frac{2}{π} u)]_{0}^{2}

= - [\frac{π}{2} § (\frac{2}{π} u)]_{0}^{2}

限度を計算する:

\frac{π}{2} § (\frac{2}{π} \cdot 2) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= - (\frac{π}{2} § (\frac{2}{π} \cdot 2) - \frac{π}{2} § (\frac{2}{π} \cdot 0))

簡素化

= - \frac{π}{2} § (2 \frac{2}{π}) + \frac{π}{2} § (0 \cdot \frac{2}{π})